hash()方法
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
高低16位先做一个异或，就是保证既能拥有低16位数据的特征，又有高16位数据的特征，相当于能让hash值更加离散。
putVal()方法
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
//tables的另一个引用
Node<K,V>[] tab;
Node<K,V> p;
int n, i;
//懒加载
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
//懒加载
//根据hashcode的高低16位异或 再和 capacity-1按位相与 得到tables[]对应的下标
//如果为空，就说明没有发生碰撞，直接占坑
//如果不为空，那么就找到了p，即对应的坑头
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
//保存对应的节点
Node<K,V> e;
K k;
//如果坑头就是目标节点
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
//如果坑头不是目标节点，且坑头是TreeNode，那么就说明链表已经变成了红黑树，要向树中继续插入或者寻找节点
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
//这种情况就是坑头不是目标节点，目标节点在链表上
else {
for (int binCount = 0; ; ++binCount) {
//如果只有坑头节点，那么就往后加一个节点
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
//节点个数大于8，就要变成红黑树
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
//变成红黑树
treeifyBin(tab, hash);
break;
}
//如果目标节点是坑头节点的后继节点
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
//继续向后遍历
p = e;
}
}
//如果目标节点已经存在
//替换为新值，返回旧值
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
//LinkedHashMap会重写这个方法，完成排序操作
afterNodeAccess(e);
return oldValue;
}
}
//fail-fast的保障机制
++modCount;
//判断是否超过了阈值
if (++size > threshold)
//扩容吧您馁
resize();
//LinkedHashMap会重写这个方法，完成排序操作
afterNodeInsertion(evict);
//插入成功，返回null
return null;
}
resize()方法
/**
* Initializes or doubles table size. If null, allocates in
* accord with initial capacity target held in field threshold.
* Otherwise, because we are using power-of-two expansion, the
* elements from each bin must either stay at same index, or move
* with a power of two offset in the new table.
*
* @return the table
*/
/**
承担着初始化，和double容量的任务
扩容结束后，要不坑位不动，要不坑位double
*/
final Node<K,V>[] resize() {
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
int oldThr = threshold;
int newCap, newThr = 0;
if (oldCap > 0) {
//如果已经扩容到最大值了，就不再扩容了
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
//如果扩容后小于最大值，那么就把阈值也double
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
}
//不太懂为什么非默认构造就要这么赋值？？？
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
else { // zero initial threshold signifies using defaults
//懒加载初始化
newCap = DEFAULT_INITIAL_CAPACITY;
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
}
if (newThr == 0) {
float ft = (float)newCap * loadFactor;
newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
(int)ft : Integer.MAX_VALUE);
}
threshold = newThr;
@SuppressWarnings({"rawtypes","unchecked"})
//申请新空间
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
//遍历旧tables，分配旧节点
if (oldTab != null) {
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
if ((e = oldTab[j]) != null) {
oldTab[j] = null;
//如果只有坑头，那就很好处理了，直接hash完塞进新tables即可
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
//如果是个红黑树节点，那就调用split对树节点完成分发
else if (e instanceof TreeNode)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
//如果是链表
else { // preserve order
Node<K,V> loHead = null, loTail = null;
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
do {
next = e.next;
//==0说明不用动，还继续放j位置就行
if ((e.hash & oldCap) == 0) {
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
}
//!=0说明之前把hash值的左边几位放弃掉了，本次要补回来，放置在j+oldCap的位置
else {
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
hiTail = e;
}
} while ((e = next) != null);
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}
remove()方法
/**
* Implements Map.remove and related methods.
*
* @param hash hash for key
* @param key the key
* @param value the value to match if matchValue, else ignored
* @param matchValue if true only remove if value is equal
* @param movable if false do not move other nodes while removing
* @return the node, or null if none
*/
final Node<K,V> removeNode(int hash, Object key, Object value,
boolean matchValue, boolean movable) {
Node<K,V>[] tab;
Node<K,V> p;
int n, index;
//必要非空检查
if ((tab = table) != null && (n = tab.length) > 0 &&
(p = tab[index = (n - 1) & hash]) != null) {
Node<K,V> node = null, e;
K k;
V v;
//寻找目标节点
//如果坑头是目标节点
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
node = p;
else if ((e = p.next) != null) {
//如果是树
if (p instanceof TreeNode)
node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
//如果是链表
else {
//遍历找目标节点
do {
if (e.hash == hash &&
((k = e.key) == key ||
(key != null && key.equals(k)))) {
node = e;
break;
}
p = e;
} while ((e = e.next) != null);
}
}
//不为空且进行必要性非空判断
if (node != null && (!matchValue || (v = node.value) == value ||
(value != null && value.equals(v)))) {
//如果是树节点
if (node instanceof TreeNode)
((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
//如果是坑头节点
else if (node == p)
tab[index] = node.next;
//如果是链表中的一个节点
else
p.next = node.next;
++modCount;
--size;
//LinkedHashMap会重写这个方法，完成排序操作
afterNodeRemoval(node);
return node;
}
}
return null;
}
get()方法
其实和remove()方法的查找过程基本一致，就是先判断key的地址是否相等，相等的话就返回true，如果不相等的话，只能代表物理地址不一致，并不代表业务含义不一致，所以还要调用equals()方法进行业务含义的判断，两者有一个相等，即可认为是相等的。
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